Evaluate `int tan^(2)x sin^(2)xdx`

Explanation

To evaluate the integral \( \int \tan^2 x \sin^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int \tan^2 x \sin^2 x \, dx \]
Recall that \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \). Therefore, we can rewrite the integral as:
\[ \int \frac{\sin^2 x}{\cos^2 x} \sin^2 x \, dx = \int \frac{\sin^4 x}{\cos^2 x} \, dx \]

### Step 2: Use the identity for \(\sin^2 x\)
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can express \(\sin^4 x\) as:
\[ \sin^4 x = (1 - \cos^2 x)^2 = 1 - 2\cos^2 x + \cos^4 x \]
Thus, the integral becomes:
\[ \int \frac{1 - 2\cos^2 x + \cos^4 x}{\cos^2 x} \, dx \]

### Step 3: Simplify the integral
Now, we can split the integral:
\[ \int \left( \frac{1}{\cos^2 x} - 2 + \cos^2 x \right) \, dx \]
This simplifies to:
\[ \int \sec^2 x \, dx - 2 \int dx + \int \cos^2 x \, dx \]

### Step 4: Evaluate each integral
1. The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x \]
2. The integral of \( -2 \) is:
\[ -2 \int dx = -2x \]
3. For \( \int \cos^2 x \, dx \), we can use the identity:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]
Thus,
\[ \int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \int dx + \frac{1}{2} \int \cos 2x \, dx \]
Evaluating this gives:
\[ = \frac{x}{2} + \frac{1}{4} \sin 2x \]

### Step 5: Combine all parts
Now, we can combine all parts:
\[ \int \tan^2 x \sin^2 x \, dx = \tan x - 2x + \left( \frac{x}{2} + \frac{1}{4} \sin 2x \right) + C \]
This simplifies to:
\[ \tan x - \frac{3x}{2} + \frac{1}{4} \sin 2x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \tan^2 x \sin^2 x \, dx = \tan x - \frac{3x}{2} + \frac{1}{4} \sin 2x + C \]